GATE CS 2012 Q.30 (CS Q.30): Boolean Algebra and Logic Minimization

What is the minimal form of the Karnaugh map shown below? Assume that $X$ denotes a don't care term.

$\overline{b}\overline{d}$

$\overline{b}\overline{d} + \overline{b}\overline{c}$

$\overline{b}\overline{d} + a\overline{b}cd$

$\overline{b}\overline{d} + \overline{b}\overline{c} + \overline{c}\overline{d}$

Official answer and solution

Solution

Group 1s and don't care conditions X on the 4-variable K-map into minimal prime implicants.

Identify the 1s and don't cares on the 4-variable K-map (columns $ab$ , rows $cd$ ):

Group 1: The four corners $(ab=00, cd=00)$ , $(ab=10, cd=00)$ , $(ab=00, cd=10)$ , and $(ab=10, cd=10)$ . In this group of 4 cells, $b = 0$ and $d = 0$ . This yields the term $\overline{b}\overline{d}$ .
Group 2: To cover the remaining 1 at $(ab=10, cd=01)$ , group cells $(ab=00, cd=00)$ , $(ab=10, cd=00)$ , $(ab=00, cd=01)$ , and $(ab=10, cd=01)$ . In these 4 cells, $b = 0$ and $c = 0$ . This yields the term $\overline{b}\overline{c}$ .

All 1s are now covered by these two groups, and don't care terms are used only to expand group sizes.
Therefore, the minimal sum of products form is $\overline{b}\overline{d} + \overline{b}\overline{c}$ , corresponding to option (B).

Final answer

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